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Find the parametric equation of the line in the \(x-y\) plane that goes through the indicated point in the direction of the indicated vector. $$ (-1,-2),\left[\begin{array}{r} 1 \\ -3 \end{array}\right] $$

Short Answer

Expert verified
Parametric equations: \\(x(t) = -1 + t\\), \\(y(t) = -2 - 3t\\)

Step by step solution

01

Point and Direction Vector

Identify the given point, which is \((-1, -2)\), and the direction vector, which is \([1, -3]\). These will be used to write the parametric equations of the line.
02

Parametric Equation Setup

Each component of the point \((-1, -2)\) and the vector \[\begin{array}{r} 1 \ -3 \ \end{array}\] will be used to form the parametric equations of the line. This gives us two separate equations, one for \((x)\) and one for \((y)\).
03

X Equation Formulation

The equation for \((x)\) is formed from the point's x-value of -1 and the x-component of the direction vector, which is 1. Therefore, the parametric equation for \((x)\) is: \( x(t) = -1 + 1t = -1 + t.\)
04

Y Equation Formulation

Similarly, the equation for \((y)\) is formed from the point's y-value of -2 and the y-component of the direction vector, which is -3. Therefore, the parametric equation for \((y)\) is: \( y(t) = -2 - 3t.\)
05

Combine Parametric Equations

Combine the two equations to have the complete set of parametric equations for the line:- \(x(t) = -1 + t\)- \(y(t) = -2 - 3t\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Direction Vector and Its Role
In the realm of parametric equations, a direction vector is fundamental. This vector determines the direction in which the line extends. Think of it as an arrow pointing the way the line stretches across the plane. In our exercise, the direction vector is \([1, -3]\). Each component of this vector affects how the corresponding coordinates change.
A direction vector, by definition, shows how much the line rises or falls, and runs left or right.
  • The first element, here being "1", indicates movement along the x-axis.
  • The second one, "-3", describes movement along the y-axis.
The increase or decrease of the parameter \(t\) in our equations shifts the position along this direction vector, literally drawing the line on the coordinate plane.
Understanding Line Equations through Parametrics
Line equations are an essential part of understanding parametric equations. Unlike traditional line equations, which are often presented in the slope-intercept form \(y = mx + c\), parametric equations separate \(x\) and \(y\) into individual expressions that depend on a common parameter \(t\).
The parametric form of a line is given by:
\(x(t) = x_0 + at\)
\(y(t) = y_0 + bt\)
Where \((x_0, y_0)\) is a point on the line, and \((a, b)\) is the direction vector.
  • This approach offers flexibility and clarity when plotting lines that aren't necessarily functions.
  • In our specific problem, the parametric equations are \(x(t) = -1 + t\) and \(y(t) = -2 - 3t\).
These equations together describe every point on the line through a free parameter \(t\), providing a dynamic way to represent and analyze lines.
Connecting Parametric Equations and Calculus Education
In calculus education, understanding parametric equations is a leap towards more advanced calculus topics. Parametric equations provide a solid foundation for curve sketching, a vital skill in calculus.
Unlike functions in the form \(y=f(x)\), parametric equations can easily deal with situations where a single x-value corresponds to multiple y-values, or where the function has no simple algebraic relationship.
This is particularly useful when describing motion or flow of multidimensional objects. For instance, consider a particle moving through space. Its path can be represented with parametric equations.
  • Understanding the derivative of these equations helps find the velocity or acceleration of the particle.
  • Grasping the integral of these equations can help determine the path length.
Thus, mastering parametric equations is an integral part of calculus education, bridging basic algebra with the dynamic analysis innate to calculus.

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Most popular questions from this chapter

33\. Let $$ A=\left[\begin{array}{rr} 2 & 1 \\ -1 & -3 \end{array}\right] $$ Find \(A^{2}, A^{3}\), and \(A^{4}\).

Suppose that breeding occurs once a year and that a census is taken at the end of each breeding season. Assume that a population is divided into three age classes and that \(20 \%\) of the females age 0 and \(70 \%\) of the females age 1 survive until the end of the next breeding season. Assume further that females age 1 have an average of \(3.2\) female offspring and females age 2 have an average of \(1.7\) female offspring. If, at time 0, the population consists of 2000 females age 0,800 females age 1, and 200 females age 2, find the Leslie matrix and the age distribution at time 2 .

Assume the given Leslie matrix \(L .\) Determine the number of age classes in the population. What fraction of two-year-olds survive until the end of the next breeding season? Determine the average number of female offspring of a one-yearold female. $$ L=\left[\begin{array}{lll} 0 & 4.2 & 3.7 \\ 0.7 & 0 & 0 \\ 0 & 0.1 & 0 \end{array}\right] $$

Suppose \(A\) is a \(3 \times 4\) matrix and \(B\) is an \(m \times n\) matrix. What are values of \(m\) and \(n\) such that the following products are defined? (a) \(A B\) (b) \(B A\)

$$ \begin{array}{r} A=\left[\begin{array}{rrr} 1 & 3 & -1 \\ 2 & 4 & 1 \\ 0 & -2 & 2 \end{array}\right], \quad B=\left[\begin{array}{rrr} 5 & -1 & 4 \\ 2 & 0 & 1 \\ 1 & -3 & -3 \end{array}\right], \\ \quad C=\left[\begin{array}{rrr} -2 & 0 & 4 \\ 1 & -3 & 1 \\ 0 & 0 & 2 \end{array}\right] \end{array} $$ $$ \text { Show that } A+B=B+A \text { . } $$

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